What is the equation of the tangent to the circle x^2 + y^2 = 25 at the point (-3, -4)?

From circle theorems, the tangent to a circle is perpendicular to the radius, and so to find the tangent we can use the gradient of the radius connecting the point (-3, -4) to the centre of the circle and then find the negative reciprocal of it...From the equation we can see that the centre of the circle is the point (0,0) as the x and y parts are written as x^2 and y^2. ( if the question had for example (x+1)^2 or (y-2)^2 this would indicate that the circle had a non origin centre point)The two points, (0,0) and (-3, -4) lie on the radius. To find the gradient of this line we use change in y/change in x...0-(-4)/0-(-3)= 4/3The tangent is perpendicular to this radius so the gradient of the tangent will be the negative reciprocal ...The negative reciprocal just means -1/your number , so in this case -1/(4/3) = -3/4Now we know that the tangents gradient we can begin to write the equation of it... it is a straight line of the form y=mx+c...y = -3/4x +cTo find the constant c, we know that the line passes through (-3, -4) so we can substitute these values into the equation...-4 = -3/4(-3) +C-4= 9/4 +C-4-9/4=CC = -25/4Finally adding C to our equation we get...Y = -3/4 X - 25/4